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=>\(\frac{7^x.\left(7^2+7+1\right)}{57}=\frac{5^{2x}.\left(1+5+5^2\right)}{131}\)
=>\(\frac{7^x.57}{57}=\frac{5^{2x}.131}{131}\)
=>7x=52x
=>7x=(52)x
=>7x=25x
=>7=25 (vô lí)
Vậy ko tìm được xthỏa mãn đề bài
1)
2x.(x-2) - x.(2x+1) = 3
=> 2x2 - 4x - 2x2 - x = 3
=> (2x2 - 2x2 ) - (4x+x) = 3
=> -5x = 3
=> x = \(\dfrac{-3}{5}\)
2) (2x-1).(x-2) - (x+3).(2x-7) = 3
=> 2x2 - 4x - x + 2 - 2x2 + 7x - 6x + 21 = 3
=> (2x2 - 2x2) - (4x + 6x + x - 7x) + 2 + 21 = 3
=> -4x = -20
=> x = -20 : (-4)
=> x = 5
3) (x - 5).(-x + 4) - (x - 1).(x + 3) = -2x2
=> Bạn tách tương tự như mấy câu 2 nhé! Nếu không làm được thì bảo mình
a/ 72+x + 2.7x-1 =345
73 . 7x-1 + 2.7x-1 = 345
7x-1 (73+2) = 345
7x-1 . 345 = 345
7x-1 = 345 : 345 = 1
7x-1 = 70
x - 1 = 0
x = 0+1 = 1
b/ 81-2x . 27x = 95
(34)-2x . (33)x = (32)5
34.(-2x) . 33.x = 32.5
3-8x . 33x = 310
3-8x+3x = 310
3-5x = 310
-5x = 10
x = 10 : (-5) = -2
a) \(\dfrac{3x-4}{2x+5}=\dfrac{3x+7}{2x-20}\left(đk:x\ne-\dfrac{5}{2},x\ne10\right)\)
\(\Rightarrow\left(3x-4\right)\left(2x-20\right)=\left(3x+7\right)\left(2x+5\right)\)
\(\Rightarrow6x^2-68x+80=6x^2+29x+35\)
\(\Rightarrow97x=45\Rightarrow x=\dfrac{45}{97}\)
b) \(\dfrac{10x-5}{7x+2}=\dfrac{50x+10}{35x-29}\left(đk:x\ne-\dfrac{2}{7},x\ne\dfrac{29}{35}\right)\)
\(\Rightarrow\left(10x-5\right)\left(35x-29\right)=\left(50x+10\right)\left(7x+2\right)\)
\(\Rightarrow350x^2-465x+145=350x^2+170x+20\)
\(\Rightarrow635x=125\Rightarrow x=\dfrac{25}{127}\)
a, 11/12 - ( 2/5 + x ) = 2/3
<=> \(\frac{2}{5}+x=\frac{11}{12}-\frac{2}{3}=\frac{1}{4}\)
=> x=\(\frac{1}{4}-\frac{11}{12}=-\frac{2}{3}\)
b, 2x . ( x - 1/7 ) = 0
<=>\(\left[\begin{array}{nghiempt}x=0\\x-\frac{1}{7}=0\end{array}\right.\)<=> \(\left[\begin{array}{nghiempt}x=0\\x=\frac{1}{7}\end{array}\right.\)
vậy x={\(0;\frac{1}{7}\)}
c, 3/4 + 1/4 : x = 2/5
<=>\(\frac{1}{4}:x=\frac{2}{5}-\frac{3}{4}=-\frac{7}{20}\)
<=> \(x=\frac{1}{4}:\left(-\frac{7}{20}\right)=-\frac{5}{7}\)
vậy x=-5/7
a) \(\frac{11}{12}-\left(\frac{2}{5}+x\right)=\frac{2}{3}\)
\(\Leftrightarrow\frac{11}{12}-\frac{2}{5}-x=\frac{2}{3}\)
\(\Leftrightarrow-x=\frac{2}{3}-\frac{11}{12}+\frac{2}{5}=\frac{3}{20}\)
\(\Leftrightarrow x=-\frac{3}{20}\)
b) \(2x\left(x-\frac{1}{7}\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x-\frac{1}{7}=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=\frac{1}{7}\end{array}\right.\)
c) \(\frac{3}{4}+\frac{1}{4}:x=\frac{2}{5}\)
\(\Leftrightarrow\frac{1}{4x}=\frac{2}{5}-\frac{3}{4}=-\frac{7}{20}\)
\(\Leftrightarrow4x=\frac{-20}{7}\)
\(\Leftrightarrow x=-\frac{5}{7}\)
\(A\left(x\right)=\left(x-2x^2\right)\left(15x^2+7\right)\)
\(A\left(x\right)=0\)\(\Leftrightarrow\left(x-2x^2\right)\left(15x^2+7\right)=0\)
\(\Leftrightarrow x-2x^2=0\Leftrightarrow x\left(1-2x\right)=0\Rightarrow\orbr{\begin{cases}x=0\\1-2x=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\2x=1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{2}\end{cases}}}\)
Hoặc \(\Leftrightarrow15x^2+7=0\Leftrightarrow15x^2=-7\Leftrightarrow x^2=\frac{-7}{15}\)(vô lí)
Vậy \(x=0,x=\frac{1}{2}\)là 2 nghiệm của \(A\left(x\right)\)
\(\left(x-2x^2\right)\left(15x^2+7\right)=0\)
Với \(x-2x^2=0\)
\(\Rightarrow x=2x^2\Rightarrow2x=1\)
\(x=\frac{1}{2}\)
Với \(15x^2+7=0\Rightarrow15x^2=-7\)
\(x^2=-\frac{7}{15}\)vô lý)
Vậy nghiệm của đa thứ A(x) là \(x=\frac{1}{2}\)
\(a,|x|=2001\)
\(\Rightarrow x=-2001;x=2001\)
\(c,3-\left(x-2\right)=-2x+7\)
\(\Rightarrow3-x+2=-2x+7\)
\(\Rightarrow5-x=-2x+7\)
\(\Rightarrow x=2\)
\(d,\left(\frac{3}{4}\right)+\frac{2}{5}x=\frac{29}{30}\)
\(\Rightarrow\frac{2}{5}x=\frac{13}{60}\)
\(\Rightarrow x=\frac{13}{24}\)
\(e,\left(\frac{3}{7}\right)^5.x=\left(\frac{3}{7}\right)^7\)
\(\Rightarrow x=\left(\frac{3}{7}\right)^2\)
2x+2-2x=96
=>2x.(22-1)=96
=>2x.3=96
=>2x=32
=>2x=25
=>x=5
7.2x=29+5.28
=>7.2x=28.(2+5)=
=>7.2x=28.7
=>2x=28
=>x=8